JEE Main 2019 (11 January)

LEVELJEE Main

Animated Solution for Mathematics - Probability: Two integers are selected at random from the set ${1, 2, \dots, 11}$ . Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

Q: Why is the sum of an odd and an even number rejected?

Think about the definition of parity. An odd number is $2k+1$ and an even number is $2m$. Their sum is $2(k+m)+1$, which is always odd. Since our condition demands an even sum, any pair containing one odd and one even number fails the test. We must discard those cases entirely.

Select Answer:

Visualized Solution

The Sample Space

Set $S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}$
Total numbers available = $11$

Categorizing Odd and Even

Odd numbers ( $n_{o} = 6$ ): ${1, 3, 5, 7, 9, 11}$
Even numbers ( $n_{e} = 5$ ): ${2, 4, 6, 8, 10}$

The Condition for an Even Sum

Condition: The sum of two selected numbers is even.
Rule 1: $Odd + Odd = Even$
Rule 2: $Even + Even = Even$
Rule 3: $Odd + Even = Odd$ (Rejected)

Case 1: Selecting Two Odd Numbers

Number of ways to select 2 odd numbers from 6:
$n (Odd, Odd) = (2 6)$

Calculating Case 1

$n (Odd, Odd) = \frac{6 \times 5}{2 \times 1}$
$n (Odd, Odd) = 15$

Case 2: Selecting Two Even Numbers

Number of ways to select 2 even numbers from 5:
$n (Even, Even) = (2 5)$

Calculating Case 2

$n (Even, Even) = \frac{5 \times 4}{2 \times 1}$
$n (Even, Even) = 10$

Total Favorable Outcomes for the Condition

Total outcomes where sum is even (New Sample Space):
$n (Sum is Even) = 15 + 10$
$n (Sum is Even) = 25$

Defining the Target Event

Target Event $A$ : Both selected numbers are even.
Favorable outcomes for $A$ : $n (Even, Even) = 10$

Setting up the Conditional Probability

$P (Both Even ∣ Sum Even) = \frac{n ( Both Even )}{n ( Sum Even )}$
$P = \frac{10}{25}$

Final Simplification

Simplifying the fraction:
$P = \frac{10 \div 5}{25 \div 5}$
$P = \frac{2}{5}$

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The Sigma Insight: Conditional Probability

Solution Diagram

Analyzing the Setup

We are working with a set of eleven distinct tokens numbered from

1

11

. This set contains six odd numbers:

{1, 3, 5, 7, 9, 11}

and five even numbers:

{2, 4, 6, 8, 10}

This parity-based classification is the fundamental structure of our sample space. We are selecting two numbers at random from this set.

The Filter of the Even Sum

The condition provided is that the sum of the two selected numbers must be even. In mathematics, parity rules dictate that:

Odd + Odd = Even

Even + Even = Even

Odd + Even = Odd

Since our condition requires an even sum, we must exclude any pair consisting of one odd and one even number. Our restricted sample space consists only of pairs that are either both odd or both even.

Counting the Possibilities

We use combinations to determine the number of ways to satisfy these conditions. For the case of selecting two odd numbers from the set of six, we calculate:

(2 6) = \frac{6 \times 5}{2 \times 1} = 15

For the case of selecting two even numbers from the set of five, we calculate:

(2 5) = \frac{5 \times 4}{2 \times 1} = 10

The total number of ways to satisfy the condition of an even sum is the sum of these two cases:

15 + 10 = 25

This value of

25

represents our new, restricted sample space.

Final Calculation

The question asks for the probability that both numbers are even, given that their sum is even. Our target event is the selection of two even numbers, for which we found

10

favorable outcomes.

The conditional probability is the ratio of these favorable outcomes to the total outcomes in our restricted sample space:

P = \frac{10}{25}

Simplifying this fraction by dividing both the numerator and the denominator by

5

, we arrive at the final result:

$P = \frac{2}{5}$

Common Traps & FAQs

Wait, why can't we just pick any two numbers from the set of 11? Why do we need to calculate the 'even sum' cases separately?

Great question! That is the trap of conditional probability. If we just picked any two numbers, we would be looking at the unconditional probability. But the problem explicitly tells us: 'Given that the sum is even.' This phrase acts like a filter, shrinking our entire universe of possibilities down to only those where the sum is even. We must respect that constraint!

Does the order in which I pick the two numbers matter? Should I use permutations instead of combinations?

In this specific problem, order does not matter. Whether you pick the number 2 then 4, or 4 then 2, the set of selected numbers is the same, and the sum is the same. Using combinations

(r n)

is the most efficient way to count sets. If you used permutations, you would just be multiplying both the numerator and denominator by

2!

, and they would cancel out anyway!

Why is the sum of an odd and an even number rejected?

Think about the definition of parity. An odd number is

2 k + 1

and an even number is

2 m

. Their sum is

2 (k + m) + 1

, which is always odd. Since our condition demands an even sum, any pair containing one odd and one even number fails the test. We must discard those cases entirely.

Animated Solution for Mathematics - Probability: Two integers are selected at random from the set ${1, 2, \dots, 11}$ . Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

Select Answer:

Visualized Solution

The Sample Space

Categorizing Odd and Even

The Condition for an Even Sum

Case 1: Selecting Two Odd Numbers

Calculating Case 1

Case 2: Selecting Two Even Numbers

Calculating Case 2

Total Favorable Outcomes for the Condition

Defining the Target Event

Setting up the Conditional Probability

Final Simplification

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Filter of the Even Sum

Counting the Possibilities

Final Calculation

Similar Questions

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A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is .........

A box contains 100 tickets numbered $1, 2, \dots, 100$ . Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability .........

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Select Answer:

Visualized Solution

The Sample Space

Categorizing Odd and Even

The Condition for an Even Sum

Case 1: Selecting Two Odd Numbers

Calculating Case 1

Case 2: Selecting Two Even Numbers

Calculating Case 2

Total Favorable Outcomes for the Condition

Defining the Target Event

Setting up the Conditional Probability

Final Simplification

The Sigma Insight: Conditional Probability

Analyzing the Setup

The Filter of the Even Sum

Counting the Possibilities

Final Calculation

Similar Questions

A die is thrown two times and the sum of the scores appearing on the die is observed to be a multiple of 4. Then the conditional probability that the score 4 has appeared atleast once is :

Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is :

A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If A and B are two events such that P(A)=31​, P(B)=51​ and P(A∪B)=21​, then P(A∣B′)+P(B∣A′) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let E1​ and E2​ be two events such that the conditional probabilities P(E1​∣E2​)=21​, P(E2​∣E1​)=43​ and P(E1​∩E2​)=81​. Then:

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One Indian and four American men and their wives are to be seated randomly around a circular table. Then the conditional probability that the Indian man is seated adjacent to his wife given that each American man is seated adjacent to his wife is

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